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Oksi-84 [34.3K]
4 years ago
7

Is muddy water a substance?

Physics
2 answers:
sergejj [24]4 years ago
7 0

Answer: No, I wouldn't imagine so.

Explanation:

Kazeer [188]4 years ago
7 0
Muddy water is not a substance
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A loop of wire carrying a steady current I is initially at rest perpendicular to a uniform magnetic field of magnitude B, as sho
seropon [69]

your answer is b to be presided

8 0
3 years ago
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When two oppositely charged electrodes are immersed in a solution, positively charged ions are attracted to the negative electro
e-lub [12.9K]

Answer:

conductivity of solution is reduced.

Explanation:

When two oppositely charged electrodes are immersed in a solution, positively charged ions are attracted to the negative electrode and gain electrons. The negatively charged ions are attracted to the positive electrode and release electrons.

Due to the process mentioned above , the negatively charged ions are accumulated at the positive electrode and the positively charged ions are accumulated at the negative electrode . This accumulation prevents further attraction of ions at oppositely charged electrodes because the incoming ions face repulsion from already accumulated ions at electrodes.  Further , it creates an emf acting in opposite direction . It reduces the current through the solution. Hence conductivity of solution is reduced.

5 0
3 years ago
How much energy is released when the correct number of protons and neutrons come together to form deuterium? a) 0.52 MeV b) 1.7
irga5000 [103]

<u>Answer:</u> The correct answer is 2.24 MeV.

<u>Explanation:</u>

The chemical reaction for the formation of deuterium from proton and neutron follows:

_{1}^1\textrm{H}+_0^1\textrm{n}\rightarrow _1^2\textrm{H}

We are given:

Mass of _{1}^{1}\textrm{H} = 1.00784 u

Mass of _{0}^{1}\textrm{n} = 1.008665 u

Mass of _{1}^{2}\textrm{H} = 2.014102 u

To calculate the mass defect, we use the equation:

\Delta m=\text{Mass of reactants}-\text{Mass of products}

\Delta m=(m_{_1^2H}+m_{n})-(m_{_1^2H})\\\\\Delta m=(1.00784+1.008665)-(2.014102)=0.002403u

To calculate the energy released, we use the equation:

E=\Delta mc^2\\E=(0.002403u)\times c^2

E=(0.002403u)\times (931.5MeV)    (Conversion factor:  1u=931.5MeV/c^2  )

E=2.24MeV

Hence, the energy released in the given nuclear reaction is 2.24 MeV.

5 0
3 years ago
The linear density rho in a rod 3 m long is 8/ x + 1 kg/m, where x is measured in meters from one end of the rod. Find the avera
Montano1993 [528]

Answer:

The average density of the rod is 1.605 kg/m.

Explanation:

The average density of the rod is given by:

\rho = \frac{m}{l}    

To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, as follows:

\int_{0}^{3} \frac{8}{3(x + 1)}dx

\rho = \frac{8}{3} \int_{0}^{3} \frac{1}{(x + 1)}dx   (1)

Using u = x+1  →  du = dx  → u₁= x₁+1 = 0+1 = 1 and u₂ = x₂+1 = 3+1 = 4

By entering the values above into (1), we have:

\rho = \frac{8}{3} \int_{0}^{3} \frac{1}{u}du

\rho = \frac{8}{3}*log(u)|_{1}^{4} = \frac{8}{3}[log(4) - log(1)] = 1.605 kg/m

Therefore, the average density of the rod is 1.605 kg/m.  

       

I hope it helps you!    

4 0
3 years ago
Similar to what you see in your textbook, you can generally omit the multiplication symbol as you answer questions online, excep
lidiya [134]

Answer:

ma= ma

m⋅a = m⋅a

And equivalently:

am=ma

a⋅m = m⋅a

Explanation:

Question

Assuming this question "Similar to what you see in your textbook, you can generally omit the multiplication symbol as you answer questions online, except when the symbol is  needed to make your meaning clear. For example, 1*10^5 is not the same as 110^5 . When you need to be explicit, type * (Shift + 8) to insert the multiplication operator. You will see a multiplication dot (⋅) appear in the answer box. Do not use the symbol x. For example, for the expression ma,

typing m⋅a would be correct, but mxa would be incorrect".

Solution to the problem

For this case we want to write a expression for ma, and based on the previous info we can write:

ma= ma

m⋅a = m⋅a

And equivalently:

am=ma

a⋅m = m⋅a

But is not correct do this:

mxa=mxa

axm = mxa

8 0
3 years ago
Read 2 more answers
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